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The problem in question is 4.

For the life of me I cannot figure out the solution, can someone explain how its done? From my attempts it appears that nothing is d-separated from A if evidence in B and E, but this is wrong.

Carlos Cinelli
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crystyxn
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  • Please type your question as text, do not just post a photograph (see here). When you retype the question, add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. – gung - Reinstate Monica Apr 11 '18 at 12:59

1 Answers1

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When not conditioning, colliding paths $\rightarrow \leftarrow$ are closed and other paths are open. When conditioning, the opposite happens: collider paths are opened (when conditioning on the collider or descendant of the collider) and the other paths are closed.

Thus, conditioning on $B$ opens the path $A \rightarrow B \leftarrow C$ and $A$ will not be separated from the variables downstream that directed path to $C, D$ and $H$.

But conditioning on $E$ closes the path $A \rightarrow E \rightarrow F$. Thus, $A$ is d-separated from $F$ and $G$, since the other paths to reach them are via colliders $D$ or $H$ which are not conditioned on.

Carlos Cinelli
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